Checking if matrix $A$ is positive definite via Cholesky decomposition How can we show that a matrix $A$ is not a positive definite matrix using the Cholesky decomposition? 
If we are not able to complete the algorithm and we cannot factor the matrix with a Cholesky decomposition does it then mean that the matrix is not positiv definite? 
Or is there an other way to check whether the matrix is positiv definite or not? 
 A: Cholesky decomposition will fail only when the matrix is not symmetric positive semi definite. Thus, if the algorithm doesn’t work, then you know your matrix is not symmetric positive semidefinite.
A: The answer to this question hinges on on the choice of arithmetic: exact or finite precision arithmetic.
In exact arithmetic arithmetic Cholesky's algorithm runs to completion and produces a lower triangular matrix $L$ with positive diagonal entries such that $A = LL^T$ if and only if $A$ is symmetric positive definite (SPD).
In finite precision arithmetic, the following two examples demonstrate that Cholesky's algorithm cannot be used to determine if a matrix is SPD. The calculations will be done in-place, i.e., I overwrite the lower triangular half of the matrices and ignore the strictly upper triangular portion. I use $\text{fl}(x)$ to denote the floating point representation of $x$. 
Example 1: A matrix $M$ which is SPD for which Cholesky's algorithm fails due to rounding errors.  Let $M$ be given by
$$ M = \begin{bmatrix} 1 + 2u & 1 \\ 1 & 1 - u \end{bmatrix}.$$
This matrix is exactly representable. It is clear that $M$ is symmetric positive definite, because the trace is
$$ \text{Tr}(M) = 2 + u > 0$$
and the determinant is
$$ \text{det}(M) = u - 2u^2 > 0.$$
It is vital understand that $\text{fl}(\sqrt{1+2u}) = 1$. We have 
$$ 1 < 1 + 2u = (1 + u)^2 - u^2 < (1+u)^2$$ which implies $$ 1 < \sqrt{1 + 2u} < 1 + u. $$ Since there are no floating point numbers in the open interval between $1$ and $1+2u$, it follows that $1$ is the floating point number which is closest to $\sqrt{1+2u}$, i.e., $\text{fl}(\sqrt{1+2u}) = 1$. 
After processing the first column of $M$ and performing the linear update of the lower right corner, we are left with
$$ M^{(1)} = \begin{bmatrix} 1  & 1 \\ 1 & - u \end{bmatrix}.$$
The algorithm will now fail because the final pivot is strictly negative.
Example 2: A symmetric matrix which has a positive and a negative eigenvalue for which Cholesky's algorithm runs to completion due to rounding errors. Let $N$ be given by
$$ N = \begin{bmatrix} 1  & 1-u \\ 1-u & 1-2u \end{bmatrix}.$$ The determinant is $$\text{det}(N) = -u^2 < 0.$$ It follows that one eigenvalue is strictly negative while the other eigenvalue is strictly positive. 
For this example it vital to understand that $$\text{fl}((1-u)^2) = 1 - 2u.$$ After processing the first column of $N$ and completing the linear update we are left with $$N^{(1)} = \begin{bmatrix} 1 & 1-u \\ 1-u & 0 \end{bmatrix}.$$ The algorithm will now run to completion because the last pivot is not negative.
In general, if Cholesky's algorithm runs to completion, then the input matrix $A$ is close to a matrix which is symmetric positive definite. Similarly, if $A$ is symmetric positive definite and if the smallest eigenvalue is sufficiently large, then Cholesky's algorithm will run to completion. The statements can be made precise, but that is perhaps a subject for another day.
 My first example is copied from my answer to a this related question. At the time, I had not constructed the second example presented here. I feel comfortable repeating the first example as they supplement each other nicely.
